正在解析 string/ISO 8601 日期

Question

我正在寻找一种将多种字符串格式解析为 dates/datetime 的方法。 as_datetime() 函数对于下面的前 3 个函数非常适合我的目的，但不是第四个。有什么建议吗？

library(lubridate)
as_datetime("20200603", tz = "America/New_York")
as_datetime("20200603231413", tz = "America/New_York")
as_datetime("20200603231413-0400", tz = "America/New_York")
as_datetime("20200528203000+0000", tz = "America/New_York")


>   as_datetime("20200603", tz = "America/New_York")
[1] "2020-06-03 EDT"

>   as_datetime("20200603231413", tz = "America/New_York")
[1] "2020-06-03 23:14:13 EDT"

>   as_datetime("20200603231413-0400", tz = "America/New_York")
Date in ISO8601 format; converting timezone from UTC to "America/New_York".
[1] "2020-06-03 23:14:13 EDT"  

>   as_datetime("20200528203000+0000", tz = "America/New_York")
[1] NA
Warning message:
All formats failed to parse. No formats found. 

>   as_datetime("20200528203116+0000", tz = "America/New_York")
[1] NA
Warning message:
All formats failed to parse. No formats found.

Answer 1

您可以使用 format= 提供更多格式，在本例中添加 %z，?strptime 定义为

     '%z' Signed offset in hours and minutes from UTC, so '-0800' is 8
          hours behind UTC. Values up to '+1400' are accepted.
          (Standard only for output.  For input R currently supports it
          on all platforms.)

library(lubridate)
as_datetime("20200603231413-0400", format = "%Y%m%d%H%M%S%z", tz = "America/New_York")
# [1] "2020-06-03 23:14:13 EDT"
as_datetime("20200628203000+0000", format = "%Y%m%d%H%M%S%z", tz = "America/New_York")
# [1] "2020-06-28 16:30:00 EDT"
as_datetime("20200528203116+0000", format = "%Y%m%d%H%M%S%z", tz = "America/New_York")
# [1] "2020-05-28 16:31:16 EDT"

正在解析 string/ISO 8601 日期

Parsing string/ISO 8601 Date

datetime

r

date

datetime-format

lubridate